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<h1 class="heading"><a href="MATH-2023-OPDE.html"><span class="title">MATH 2023: Ordinary and Partial Differential Equations</span></a></h1>
<p class="byline">Xiaoyi Chen and Wei Zhang</p>
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<a href="ch_first.html" data-scroll="ch_first" class="internal"><span class="codenumber">1</span> <span class="title">Introduction</span></a><ul>
<li><a href="sec_1-intro.html" data-scroll="sec_1-intro" class="internal">Classification of Differential Equations</a></li>
<li><a href="sec_2-intro.html" data-scroll="sec_2-intro" class="internal">Linear and Nonlinear Equation</a></li>
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<a href="ch_second.html" data-scroll="ch_second" class="internal"><span class="codenumber">2</span> <span class="title">First Order Ordinary Differential Equations</span></a><ul>
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<li><a href="sec2_2.html" data-scroll="sec2_2" class="internal">Further Discussion of Linear Equations (For reading only)</a></li>
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<li><a href="sec3_1.html" data-scroll="sec3_1" class="internal">Homogeneous equations with constant coefficient</a></li>
<li><a href="sec3_2.html" data-scroll="sec3_2" class="internal">Fundamental Solutions of Linear Homogeneous Equations</a></li>
<li><a href="sec3_3.html" data-scroll="sec3_3" class="internal">Linear Independence and Wronskian</a></li>
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<a href="ch_five.html" data-scroll="ch_five" class="internal"><span class="codenumber">5</span> <span class="title">Series Solutions of Second Order Linear Equations</span></a><ul>
<li><a href="sec5_1.html" data-scroll="sec5_1" class="internal">Brief Review on Power Series</a></li>
<li><a href="sec5_2.html" data-scroll="sec5_2" class="internal">Introduction</a></li>
<li><a href="sec5_3.html" data-scroll="sec5_3" class="active">Series Solutions Near an Ordinary Point</a></li>
<li><a href="sec5_4.html" data-scroll="sec5_4" class="internal">Euler’s Equation</a></li>
<li><a href="sec5_5.html" data-scroll="sec5_5" class="internal">Series Solution near a Regular Singular Point</a></li>
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<a href="ch_six.html" data-scroll="ch_six" class="internal"><span class="codenumber">6</span> <span class="title">System of First Order Linear Equations</span></a><ul>
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<a href="ch_seven.html" data-scroll="ch_seven" class="internal"><span class="codenumber">7</span> <span class="title">Partial Differential Equations</span></a><ul>
<li><a href="sec7_1.html" data-scroll="sec7_1" class="internal">Two-Point Boundary Value Problems</a></li>
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<li><a href="sec7_5.html" data-scroll="sec7_5" class="internal">Even and Odd Functions</a></li>
<li><a href="sec7_6.html" data-scroll="sec7_6" class="internal">Introduction to Partial Differential Equations</a></li>
<li><a href="sec7_7.html" data-scroll="sec7_7" class="internal">1D Heat Equation; Solutions by Separation of Variable and Fourier Series</a></li>
<li><a href="sec7_8.html" data-scroll="sec7_8" class="internal">Other Heat Conduction Problems</a></li>
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<a href="ch_eight.html" data-scroll="ch_eight" class="internal"><span class="codenumber">8</span> <span class="title">Laplace transform</span></a><ul>
<li><a href="sec8_1.html" data-scroll="sec8_1" class="internal">What are Laplace Transforms, and Why?</a></li>
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<li><a href="sec8_4.html" data-scroll="sec8_4" class="internal">Solving ODEs and ODE Systems</a></li>
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<li><a href="sec8_6.html" data-scroll="sec8_6" class="internal">Laplace transform for PDE (heat equation)</a></li>
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<li class="link"><a href="solutions-1.html" data-scroll="solutions-1" class="internal"><span class="codenumber">A</span> <span class="title">Selected Hints</span></a></li>
<li class="link"><a href="solutions-2.html" data-scroll="solutions-2" class="internal"><span class="codenumber">B</span> <span class="title">Selected Solutions</span></a></li>
<li class="link"><a href="appendix-1.html" data-scroll="appendix-1" class="internal"><span class="codenumber">C</span> <span class="title">List of Symbols</span></a></li>
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<main class="main"><div id="content" class="pretext-content"><section class="section" id="sec5_3"><h2 class="heading hide-type">
<span class="type">Section</span> <span class="codenumber">5.3</span> <span class="title">Series Solutions Near an Ordinary Point</span>
</h2>
<p id="p-213"><dfn class="terminology">Theorem</dfn> If <span class="process-math">\(x=x_0\)</span> is an ordinary point of</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
P(x) y^{\prime \prime}+Q(x) y^{\prime}+R(x) y=0,
\end{equation*}
</div>
<p class="continuation">then the general solution of this equation is given by</p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq5_4">
\begin{equation}
y=\sum_{n=0}^{\infty} a_n (x-x_0)^n=a_0 y_1(x)+a_1 y_2(x),\tag{5.3.1}
\end{equation}
</div>
<p class="continuation">where <span class="process-math">\(a_0\)</span> and <span class="process-math">\(a_1\)</span> are arbitrary constants and <span class="process-math">\(y_1(x)\)</span> and <span class="process-math">\(y_2(x)\)</span> are two linear independent solutions. Further, the radius of convergence for each of series solutions <span class="process-math">\(y_1\)</span> and <span class="process-math">\(y_2\)</span> is at least as large as the minimum of the radius of convergence of the series for <span class="process-math">\(Q(x)/P(x)\)</span> and <span class="process-math">\(R(x)/P(x)\text{.}\)</span></p>
<p id="p-214"><dfn class="terminology">Example 1</dfn>Determine a lower bound for the radius of convergence of series solutions about <span class="process-math">\(x=4\)</span> for</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
(x^2-2x-3) y^{\prime \prime}+x y^{\prime}+4 y=0.
\end{equation*}
</div>
<p id="p-215"><dfn class="terminology">Solution:</dfn></p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
\frac{Q(x)}{P(x)}&amp;=\frac{x}{x^2-2x-3}=\frac{x}{(x-3)(x+1)}=\frac{1}{4}\left[\frac{3}{x-3}+\frac{1}{x+1}   \right].
%&amp;=\frac{1}{4} \left[ \frac{3}{1+x-4} +\frac{1}{5+x-4}\right].\\
%\textrm{Since}~ &amp;\frac{3}{1+x-4}=\frac{3}{1+t}=3 [1-t+t^2-t^3+\cdots],\quad |t|&lt;1,\\
%&amp;\qquad \qquad=3 [1-(x-4)+(x-4)^2-(x-4)^3+\cdots],\quad |x-4|&lt;1.
\end{aligned}
\end{equation*}
</div>
<div class="displaymath process-math" data-contains-math-knowls="" id="p-216">
\begin{equation*}
\frac{3}{x-3}=\frac{3}{1+x-4}=\frac{3}{1+t}=3 [1-t+t^2-t^3+\cdots]=3 [1-(x-4)+(x-4)^2-(x-4)^3+\cdots],
\end{equation*}
</div>
<p class="continuation">where <span class="process-math">\(t=x-4\)</span> and the series is convergent for <span class="process-math">\(|t|&lt;1\)</span> or <span class="process-math">\(|x-4|&lt;1\text{.}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
\frac{1}{x+1}=\frac{1}{5+(x-4)}&amp;=\frac{1}{5}\frac{1}{1+\frac{x-4}{5}}=\frac{1}{5}[1-t+t^2-t^3+\cdots]\\
&amp;=\frac{1}{5} \left[ 1-\frac{x-4}{5}+\left(\frac{x-4}{5} \right)^2 -\left(\frac{x-4}{5}\right)^3 +\cdots  \right],
\end{aligned}
\end{equation*}
</div>
<p class="continuation">where <span class="process-math">\(t=(x-4)/5\)</span> and the series is convergent for <span class="process-math">\(\left| \frac{x-4}{5} \right|&lt;1\)</span> or <span class="process-math">\(|x-4|&lt;5\text{.}\)</span> Therefore, <span class="process-math">\(\frac{Q(x)}{P(x)}\)</span> is convergent for <span class="process-math">\(|x-4|&lt;1\)</span> and the radius of convergence is <span class="process-math">\(1\text{.}\)</span> Similarly,</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\frac{R(x)}{P(x)}=\frac{4}{(x-3)(x+1)}=\frac{1}{x-3}-\frac{1}{x+1},
\end{equation*}
</div>
<p class="continuation">which is convergent for <span class="process-math">\(|x-4|&lt;1\)</span> and the radius of convergence is <span class="process-math">\(1\text{.}\)</span> According to the theorem, a lower bound for the radius of convergence for the series solution is <span class="process-math">\(1\text{.}\)</span></p>
<p id="p-217"><dfn class="terminology">Example 2</dfn>Use the method of power series to find two linear independent series solutions about <span class="process-math">\(x=2\)</span> (first four non-zero terms only) for</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
2 y^{\prime \prime}+(x+1) y^{\prime}+3 y=0.
\end{equation*}
</div>
<p id="p-218"><dfn class="terminology">Solution:</dfn> <span class="process-math">\(x=2\)</span> is an ordinary point. According to the theorem, we have the following general solution:</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_5.html">
\begin{equation*}
y=\sum_{n=0}^{\infty} a_n (x-2)^n=a_0 y_1(x)+a_1 y_2(x).
\end{equation*}
</div>
<p class="continuation">Substituting it into the ODE,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_5.html">
\begin{equation*}
\begin{aligned}
&amp; 2 \sum_{n=0}^{\infty} n (n-1) a_n (x-2)^{n-2}+(x-2+3) \sum_{n=0}^{\infty}n  a_n (x-2)^{n-1}+3 \sum_{n=0}^{\infty} a_n (x-2)^n=0,\\
&amp;\to 2 \sum_{k=-2}^{\infty} (k+2) (k+1) a_{k+2} (x-2)^k+\sum_{n=0}^{\infty} n a_n (x-2)^n+3 \sum_{n=0}^{\infty} n a_{n} (x-2)^{n-1}+3 \sum_{n=0}^{\infty} a_n (x-2)^n=0,\\
&amp;\to 2 \sum_{n=0} (n+2) (n+1) a_{n+2} (x-2)^n++\sum_{n=0}^{\infty} n a_n (x-2)^n+3 \sum_{n=0}^{\infty} (n+1) a_{n+1} (x-2)^n+3 \sum_{n=0}^{\infty} a_n (x-2)^n=0,\\
&amp;\to \sum_{n=0}^{\infty} \left[    2 (n+2)(n+1) a_{n+2} +3(n+1) a_{n+1}+(n+3) a_n
\right] (x-2)^n=0,
\end{aligned}
\end{equation*}
</div>
<p class="continuation">which finally gives</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_5.html" id="eq5_5">
\begin{equation}
2(n+2)(n+1) a_{n+2}+3(n+1) a_{n+1}+(n+3) a_n=0.\tag{5.3.2}
\end{equation}
</div>
<p class="continuation">According to the theorem,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_5.html">
\begin{equation*}
y_1(x)=\sum_{n=0}^{\infty} a_n (x-2)^n,
\end{equation*}
</div>
<p class="continuation">with <span class="process-math">\(a_1=0, a_0=1\text{.}\)</span> In (<a href="" class="xref" data-knowl="./knowl/eq5_5.html" title="Equation 5.3.2">(5.3.2)</a>), let <span class="process-math">\(a_1=0, a_0=1\text{.}\)</span> Then</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_5.html">
\begin{equation*}
\begin{aligned}
&amp; n=0: 2 \cdot 2 \cdot 1\cdot  a_2+3 \cdot 1 \cdot 0 +3 \cdot 1=0 \to a_2=-\frac{3}{4},\\
&amp;n=1: 2 \cdot 3 \cdot 2 \cdot a_3+3 \cdot 2 \cdot a_2+4 \cdot 0=0 \to a_3=-\frac{1}{2} a_2=\frac{3}{8},\\
&amp;n=2: 2 \cdot 4 \cdot 3 \cdot a_4+3 \cdot 3 \cdot \frac{3}{8}+6\cdot \left(-\frac{3}{4}\right)=0 \to a_4=\frac{3}{64}.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Thus,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_5.html">
\begin{equation*}
y_1(x)=1-\frac{3}{4}(x-2)^2+\frac{3}{8}(x-2)^3+\frac{3}{64}(x-2)^4+\cdots
\end{equation*}
</div>
<p id="p-219">According to the theorem,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_5.html">
\begin{equation*}
y_2(x)=\sum_{n=0}^{\infty} a_n (x-2)^n,
\end{equation*}
</div>
<p class="continuation">with <span class="process-math">\(a_0=0,a_1=1.\)</span> In (<a href="" class="xref" data-knowl="./knowl/eq5_5.html" title="Equation 5.3.2">(5.3.2)</a>), let <span class="process-math">\(a_0=0, a_1=1\text{.}\)</span> Then</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_5.html">
\begin{equation*}
\begin{aligned}
&amp;n=0: 2 \cdot 2 \cdot 1 \cdot a_2+3 \cdot 1 \cdot a_1=0 \to a_2=-\frac{3}{4},\\
&amp;n=1: 2 \cdot 3 \cdot 2 \cdot a_3+ 3 \cdot 2 \cdot \left(-\frac{3}{4}\right)+4 \cdot 1=0 \to a_3=\frac{1}{24},\\
&amp;n=2: 2 \cdot 4 \cdot 3 \cdot a_4+3 \cdot 3 \cdot \frac{1}{24}+5 \cdot \left(-\frac{3}{4}\right)=0 \to a_4=\frac{9}{64}.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Thus,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_5.html">
\begin{equation*}
y_2(x)=(x-2)-\frac{3}{4}(x-2)^2+\frac{1}{24} (x-2)^3+
\frac{9}{64}(x-4)^4+\cdots.
\end{equation*}
</div>
<p id="p-220"><dfn class="terminology">Example 3</dfn> (The Legendre Equation)Find two linear independent solutions for</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
(1-x^2) y^{\prime \prime}-2 x y^{\prime}+\alpha (\alpha+1) y=0, \quad (\alpha &gt;-1)
\end{equation*}
</div>
<p class="continuation">in terms of power series about <span class="process-math">\(x=0\)</span> for <span class="process-math">\(|x|&lt;1\text{.}\)</span></p>
<p id="p-221"><dfn class="terminology">Solution:</dfn> <span class="process-math">\(x=0\)</span> is an ordinary point. According to the theorem, the general solution is given by</p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq5_6">
\begin{equation}
y=\sum_{n=0}^{\infty} a_n x^n=a_0 y_1(x)+a_1 y_2(x).\tag{5.3.3}
\end{equation}
</div>
<p class="continuation">Substituting the solution into the ODE:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp;(1-x^2) \sum_{n=0}^{\infty} n(n-1) a_n x^{n-2}-2 x \sum_{n=0}^{\infty} n a_n x^{n-1}+\alpha (\alpha+1) \sum_{n=0}^{\infty} a_n x^n=0,\\
&amp;\to \sum_{n=0}^{\infty} n(n-1) a_n x^{n-2}-\sum_{n=0}^{\infty} n(n-1) a_n x^n-2 \sum_{n=0}^{\infty} n a_n x^n+\alpha (\alpha+1) \sum_{n=0}^{\infty} a_n x^n=0,\\
&amp;\to \sum_{k=-2}^{\infty} (k+2)(k+1) a_{k+2} x^k+\sum_{n=0}^{\infty} [-n(n-1)-2 n+\alpha (\alpha+1)] a_n x^n=0,\\
&amp;\to \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n+\sum_{n=0}^{\infty} [-n(n-1)-2 n+\alpha (\alpha+1)] a_n x^n=0,\\
&amp;\to \sum_{n=0}^{\infty} \left[(n+2)(n+1) a_{n+2}+\left(-n(n-1)-2 n+\alpha (\alpha+1)\right) a_n \right] x^n=0,
\end{aligned}
\end{equation*}
</div>
<p class="continuation">which gives the recurrence relation as</p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq5_7">
\begin{equation}
a_{n+2}=\frac{-n(n+1)+\alpha (\alpha+1)}{-(n+2)(n+1)} a_n=-\frac{(\alpha-n)(\alpha+n+1)}{(n+2)(n+1)}a_n.\tag{5.3.4}
\end{equation}
</div>
<p id="p-222">From (<a href="" class="xref" data-knowl="./knowl/eq5_6.html" title="Equation 5.3.3">(5.3.3)</a>):</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_6.html ./knowl/eq5_7.html">
\begin{equation*}
y_1(x)=\sum_{n=0}^{\infty} a_n x^n, \quad \textrm{with}~ a_0=1, a_1=0.
\end{equation*}
</div>
<p class="continuation">In (<a href="" class="xref" data-knowl="./knowl/eq5_7.html" title="Equation 5.3.4">(5.3.4)</a>), let <span class="process-math">\(a_0=1, a_1=0\text{.}\)</span> Then</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_6.html ./knowl/eq5_7.html">
\begin{equation*}
\begin{aligned}
n=0: &amp;\quad a_2=-\frac{\alpha (\alpha+1)}{2!}, \quad (k=1)\\
n=1: &amp; \quad a_3=0,\\
n=2: &amp;\quad a_4=-\frac{(\alpha-2) (\alpha+3)}{4 \cdot 3} a_2=\frac{\alpha (\alpha-2) (\alpha+1) (\alpha+3)}{4!}, \quad (k=2)\\
n=3: &amp;\quad a_5=0,\\
n=4: &amp;\quad a_6=-\frac{(\alpha-4) (\alpha+5)}{6 \cdot 5} a_4=-\frac{\alpha (\alpha-2) (\alpha-4) (\alpha+1) (\alpha+3) (\alpha+5)}{6!},\quad (k=3)\\
&amp;\quad \cdots\\
&amp;\quad a_{2 k+1}=0,\\
&amp; \quad a_{2 k}=(-1)^k \frac{\alpha (\alpha-2) (\alpha-4) \cdots (\alpha-2 (k-1)) (\alpha+1) (\alpha+3) (\alpha+5) \cdots (\alpha+2 k-1)}{(2 k)!}, \quad k \geq 1.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Thus,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_6.html ./knowl/eq5_7.html">
\begin{equation*}
\begin{aligned}
y_1=\sum_{n=0}^{\infty} a_n x^n&amp;=1+\sum_{k=1}^{\infty} a_{2 k} x^{2 k}+\sum_{k=0}^{\infty} a_{2k+1} x^{2k+1}\\
&amp;=1+\sum_{k=1}^{\infty} (-1)^k \frac{\alpha (\alpha-2) \cdots (\alpha-2k+2) (\alpha+1)(\alpha+3)\cdots (\alpha+2k-1)}{(2k)!} x^{2k}.
\end{aligned}
\end{equation*}
</div>
<p id="p-223">From (<a href="" class="xref" data-knowl="./knowl/eq5_6.html" title="Equation 5.3.3">(5.3.3)</a>),</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_6.html ./knowl/eq5_7.html">
\begin{equation*}
y_2(x)=\sum_{n=0}^{\infty} a_n x^n,\quad \textrm{with}~a_0=0, a_1=1.
\end{equation*}
</div>
<p class="continuation">In (<a href="" class="xref" data-knowl="./knowl/eq5_7.html" title="Equation 5.3.4">(5.3.4)</a>), let <span class="process-math">\(a_0=0, a_1=1\text{.}\)</span> Then</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_6.html ./knowl/eq5_7.html">
\begin{equation*}
\begin{aligned}
n=0:&amp;\quad a_2=0,\\
n=1: &amp;\quad a_3=-\frac{(\alpha-1)(\alpha+2)}{3!},\quad (k=1)\\
n=2: &amp;\quad a_4=0,\\
n=3: &amp;\quad a_5=-\frac{(\alpha-3)(\alpha+4)}{5 \cdot 4}a_3=\frac{(\alpha-1)(\alpha-3)(\alpha+2)(\alpha+4)}{5!}, \quad (k=2)\\
n=4: &amp;\quad a_6=0,\\
n=5: &amp; \quad a_7=-\frac{(\alpha-5)(\alpha+6)}{7\cdot 6}=-\frac{(\alpha-1)(\alpha-3)(\alpha-5)(\alpha+2)(\alpha+4)(\alpha+6)}{7!},\quad (k=3)\\
&amp;\quad \cdots\\
&amp;\quad a_{2k}=0,\\
&amp;\quad a_{2k+1}=(-1)^k \frac{(\alpha-1)(\alpha-3)(\alpha-5)\cdots(\alpha-2k+1)(\alpha+2)(\alpha+4)\cdots(\alpha+2k)}{(2k+1)!}, \quad k\geq1.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Thus,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_6.html ./knowl/eq5_7.html">
\begin{equation*}
\begin{aligned}
y_2=\sum_{n=0}^{\infty} a_n x^n=x+\sum_{n=2}^{\infty} a_n x^n=&amp;x+\sum_{k=1}^{\infty} a_{2k} x^{2k}+\sum_{k=1}^{\infty} a_{2k+1} x^{2k+1}\\
&amp;=x+\sum_{k=1}^{\infty} (-1)^k \frac{(\alpha-1)(\alpha-3)\cdots(\alpha-2k+1)(\alpha+2)(\alpha+4)\cdots(\alpha+2k)}{(2 k+1)!} x^{2k+1}.
\end{aligned}
\end{equation*}
</div>
<p id="p-224">Note: Consider the case that <span class="process-math">\(\alpha\)</span> is a positive integer.(i) For <span class="process-math">\(\alpha=2 l,\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y_1=1+\sum_{k=1}^{\infty} (-1)^k \frac{\alpha (\alpha-2) \cdots (\alpha-2k+2) (\alpha+1)(\alpha+3)\cdots (\alpha+2k-1)}{(2k)!} x^{2k}.
\end{equation*}
</div>
<p class="continuation">Consider <span class="process-math">\(k=l+1\text{,}\)</span> then <span class="process-math">\(\alpha-2k+2=2 l - 2l=0\text{.}\)</span> For <span class="process-math">\(k \geq l+1\text{,}\)</span> all the coefficients will be zero. Therefore,</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y_1=1+\sum_{k=1}^{l} (-1)^k \frac{\alpha (\alpha-2) \cdots (\alpha-2k+2) (\alpha+1)(\alpha+3)\cdots (\alpha+2k-1)}{(2k)!} x^{2k},
\end{equation*}
</div>
<p class="continuation">which is a polynomial of degree <span class="process-math">\(2l\text{.}\)</span>(ii) For <span class="process-math">\(\alpha=2l+1\text{,}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y_2=x+\sum_{k=1}^{\infty} (-1)^k \frac{(\alpha-1)(\alpha-3)\cdots(\alpha-2k+1)(\alpha+2)(\alpha+4)\cdots(\alpha+2k)}{(2 k+1)!} x^{2k+1}.
\end{equation*}
</div>
<p class="continuation">Consider <span class="process-math">\(k=l+1\text{,}\)</span> <span class="process-math">\(\alpha-2k+1=0\text{.}\)</span> For <span class="process-math">\(k \geq l+1\text{,}\)</span> all the coefficients will be zero. Therefore,</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y_2=x+\sum_{k=1}^{l} (-1)^k \frac{(\alpha-1)(\alpha-3)\cdots(\alpha-2k+1)(\alpha+2)(\alpha+4)\cdots(\alpha+2k)}{(2 k+1)!} x^{2k+1},
\end{equation*}
</div>
<p class="continuation">which is a polynomial of degree <span class="process-math">\(2l+1\text{.}\)</span></p></section></div></main>
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